Question

$PQR$ is a triangular park with $PQ=PR=200m$. A T.V. tower stands at the mid-point of $QR$. If the angles of elevation of the top of the tower at $P,Q$ and R are respectively $45°,30°\text{and}30°,$ then the height of the tower (in m) is :

• A. $100\sqrt{3}$
• B. $50\sqrt{2}$
• C. $100$
• D. $50$

Answer 1

The correct option is C

$100$

Explanation for the correct option.

Finding the height of the tower (in m)

Given:$PQR$ is a triangular park with $PQ=PR=200m$. A T.V. tower stands at the mid-point of $QR$. If the angles of elevation of the top of the tower at $P,Q$ and R are respectively $45°,30°\text{and}30°$

Let the height of the tower $MN$ be h

In $\Delta QMN\tan 30°=\frac{MN}{QM}$

$\therefore QM=3h=MR\left[\because M\text{is the point}QM=MR\right]...\left(1\right)$

Now in $\Delta MNP$

$\because \angle MPN=\angle NMP=45°$

$$\begin{gathered} \therefore MN=PM[\because \text{in a triangle sides opposite to equal angles are equal}]...\left(2\right) \\ \therefore MN=PM=h \end{gathered}$$

In $\Delta PMQ$

By using Equation $\left(1\right)$

$PM=\sqrt{{\left(200\right)}^2-{(\sqrt{3}h)}^2}$ [By using Pythagoras theorem]

Therefore From $\left(2\right)$

$$\begin{gathered} h=\sqrt{{\left(200\right)}^2-{(\sqrt{3}h)}^2} \\ \Rightarrow h=\sqrt{\left(40000\right)-3h^2} \\ \Rightarrow h^2=40000-3h^2\left[\text{Bysquaringboththesides}\right] \\ \Rightarrow 4h^2=40000 \\ \Rightarrow h^2=10000m \\ \Rightarrow h=100m \end{gathered}$$

Hence the correct answer is option(C)